\(\int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx\) [662]

Optimal result

Integrand size = 18, antiderivative size = 10 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \arctan \left (e^{x^2}\right ) \]

[Out]

1/2*arctan(exp(x^2))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6847, 2281, 209} \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \arctan \left (e^{x^2}\right ) \]

[In]

Int[(E^x^2*x)/(1 + E^(2*x^2)),x]

[Out]

ArcTan[E^x^2]/2

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {e^x}{1+e^{2 x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{x^2}\right ) \\ & = \frac {1}{2} \tan ^{-1}\left (e^{x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \arctan \left (e^{x^2}\right ) \]

[In]

Integrate[(E^x^2*x)/(1 + E^(2*x^2)),x]

[Out]

ArcTan[E^x^2]/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\arctan \left ({\mathrm e}^{x^{2}}\right )}{2}\)	\(8\)
default	\(\frac {\arctan \left ({\mathrm e}^{x^{2}}\right )}{2}\)	\(8\)
risch	\(\frac {i \ln \left ({\mathrm e}^{x^{2}}+i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x^{2}}-i\right )}{4}\)	\(24\)

[In]

int(exp(x^2)*x/(1+exp(2*x^2)),x,method=_RETURNVERBOSE)

[Out]

1/2*arctan(exp(x^2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \, \arctan \left (e^{\left (x^{2}\right )}\right ) \]

[In]

integrate(exp(x^2)*x/(1+exp(2*x^2)),x, algorithm="fricas")

[Out]

1/2*arctan(e^(x^2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (7) = 14\).

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.70 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\operatorname {RootSum} {\left (16 z^{2} + 1, \left ( i \mapsto i \log {\left (4 i + e^{x^{2}} \right )} \right )\right )} \]

[In]

integrate(exp(x**2)*x/(1+exp(2*x**2)),x)

[Out]

RootSum(16*_z**2 + 1, Lambda(_i, _i*log(4*_i + exp(x**2))))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \, \arctan \left (e^{\left (x^{2}\right )}\right ) \]

[In]

integrate(exp(x^2)*x/(1+exp(2*x^2)),x, algorithm="maxima")

[Out]

1/2*arctan(e^(x^2))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {1}{2} \, \arctan \left (e^{\left (x^{2}\right )}\right ) \]

[In]

integrate(exp(x^2)*x/(1+exp(2*x^2)),x, algorithm="giac")

[Out]

1/2*arctan(e^(x^2))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {e^{x^2} x}{1+e^{2 x^2}} \, dx=\frac {\mathrm {atan}\left ({\mathrm {e}}^{x^2}\right )}{2} \]

[In]

int((x*exp(x^2))/(exp(2*x^2) + 1),x)

[Out]

atan(exp(x^2))/2